Question

A microscope consists of an objective of focal length $1.9cm$ and eye piece of focal length $5cm$. The two lenses are kept at a distance of $10.5cm$. If the image is to be formed at the least distance of distinct vision, the distance at which the object is to be placed before the objective is:
(Least distance of distinct vision is 25 cm)

• A. $6.2cm$
• B. $2.7cm$
• C. $21.0cm$
• D. $4.17cm$

Answer 1

Answer: (B)

$2.7cm$

Let us find the position of first image,

$-\frac{1}{25}-\frac{1}{u}=\frac{1}{5}$.

Hence, $u=-\frac{25}{6}cm$.

Hence, image distance for first refraction (through objective) is $10.5-\frac{25}{6}=\frac{38}{6}cm$.

Let position of object be u.

Then, $\frac{6}{38}-\frac{1}{u}=\frac{1}{1.9}$

Hence, $u=-2.7cm$